CODEWITHSUNDEEP
c++inheritancetemplates
Surya
Surya

Reputation: 1197

Function template specialization in derived class

I've a base class with a function template.

I derive from base class and try to have a specialization for the function template in derived class

I did something like this.

class Base 
{
..
template <typename T>
fun (T arg) { ... }

};

class Derived : public Base
{
...
} ;

template <>
Derived::fun(int arg);

and in .cpp file I've provided implementation for the template specialization.

This works fine with MSVC 8.0 and g++-4.4.2 complains about lack of function declaration fun in Derived class.

I do not know which compiler is behaving correctly. Any help in this is greatly appreciated.

Thanks in advance, Surya

Upvotes: 12

Views: 11402

Answers (4)

Leandro T. C. Melo
Leandro T. C. Melo

Reputation: 4032

Also, an alternative option would be a plain non-template function in Derived...


class Derived : public Base
{
public:
  void fun(int) { /* ... */ }
};

Upvotes: 0

WolfgangP
WolfgangP

Reputation: 3233

g++ behaves correctly, because fun is defined in Base.

Upvotes: 0

Adam Bowen
Adam Bowen

Reputation: 11240

You need to declare the function in Derived in order to be able to overload it:

class Derived : public Base
{
    template <typename T>
    void fun (T arg) 
    {
        Base::fun<T>(arg);
    }

} ;

template <>
void Derived::fun<int>(int arg)
{
    // ...
}

Note that you may need to inline the specialisation or move it to an implementation file, in which case you must prototype the specialisation in the header file as:

template <>
void Derived::fun<int>(int arg);

otherwise the compiler will use the generalised version of 'fun' to generate code when it is called instead of linking to the specialisation.

Upvotes: 7

Andreas Brinck
Andreas Brinck

Reputation: 52549

Why can't you do

template <>
Base::fun(int arg);

g++'s error message looks right to me. fun is declared in Base and not in Derived.

Upvotes: 1

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